Problem 32 - The Calculus with Analytic Geometry

Question

Water is flowing into a tank in the form of a right-circular cylinder at the rate of \(\frac{4}{5} \pi \mathrm{ft}^{3} / \mathrm{min}\). The tank is stretching in such a way that even though it remains cylindrical, its radius is increasing at the rate of \(0.002 \mathrm{ft} / \mathrm{min}\). How fast is the surface of the water rising when the radius is \(2 \mathrm{ft}\) and the volume of water in the tank is \(20 \pi \mathrm{ft}^{3}\) ?

Step-by-Step Solution

Answer

The surface of the water is rising at \( 0.19 \ \text{ft}/\text{min} \).

Step 1 - Identify the given values

Given:- Rate of flow of water into the tank: \ \( \frac{4}{5} \pi \ \text{ft}^3/\text{min} \) - Rate of increase in radius: \ \( \frac{dr}{dt} = 0.002 \ \text{ft}/\text{min} \) - Radius when measured: \ \( r = 2 \ \text{ft} \) - Volume of water: \ \( V = 20 \pi \ \text{ft}^3 \)

Step 2 - Write the formula for the volume of a cylinder

The volume of a right circular cylinder is given by\[ V = \pi r^2 h \]where \( V \) is the volume, \( r \) is the radius, and \( h \) is the height.

Step 3 - Differentiate the volume formula with respect to time

To find how fast the surface of the water is rising, we need to differentiate the volume formula with respect to time (\( t \)):\[ \frac{dV}{dt} = \pi (2r \frac{dr}{dt}h + r^2 \frac{dh}{dt}) \]

Step 4 - Substitute given values into the differentiated formula

We know \( \frac{dV}{dt} = \frac{4}{5}\pi \ \text{ft}^3/\text{min} \), \( r = 2 \ \text{ft} \), and \( \frac{dr}{dt} = 0.002 \ \text{ft}/\text{min} \):\[ \frac{4}{5}\pi = \pi (2 \cdot 2 \cdot 0.002 \cdot h + 2^2 \cdot \frac{dh}{dt}) \]

Step 5 - Simplify and solve for \( \frac{dh}{dt} \)

Simplify the equation to find the rate at which the surface of the water is rising (\( \frac{dh}{dt} \)):\[ \frac{4}{5}\pi = \pi (0.008h + 4 \frac{dh}{dt}) \]Divide by \( \pi \):\[ \frac{4}{5} = 0.008h + 4 \frac{dh}{dt} \]

Step 6 - Determine height \( h \) when radius \( r = 2 \ \text{ft} \)

From the volume formula \( V = \pi r^2 h \), given \( V = 20\pi \ \text{ft}^3 \) and \( r = 2 \ \text{ft} \):\[ 20\pi = \pi (2)^2 h \]Solve for \( h \):\[ 20 = 4h \]\[ h = 5 \ \text{ft} \]

Step 7 - Substitute \( h \) back into the equation

Substitute \( h = 5 \ \text{ft} \) back into the equation from Step 5:\[ \frac{4}{5} = 0.008 \cdot 5 + 4 \frac{dh}{dt} \]\[ \frac{4}{5} = 0.04 + 4 \frac{dh}{dt} \]

Step 8 - Solve for \( \frac{dh}{dt} \)

Isolate \( \frac{dh}{dt} \):\[ \frac{4}{5} - 0.04 = 4 \frac{dh}{dt} \]\[ 0.8 - 0.04 = 4 \frac{dh}{dt} \]\[ 0.76 = 4 \frac{dh}{dt} \]\[ \frac{dh}{dt} = \frac{0.76}{4} \]\[ \frac{dh}{dt} = 0.19 \ \text{ft}/\text{min} \]

Key Concepts

Calculus

In this exercise, calculus plays a crucial role. Calculus is the branch of mathematics that deals with continuous change. It allows us to understand various rates of change within different contexts. Here, we use calculus to study how the height of the water in a cylindrical tank changes as water flows in. Calculus concepts like differentiation and related rates help us solve for unknowns by understanding how different variables in a scenario are changing over time. These concepts are essential for solving real-world problems involving motion, growth, and other dynamic processes.
When dealing with physical problems involving rates of change, we generally start with known formulas and relationships, such as the volume of a cylinder. Through calculus, we then find how one variable influences another.

Related Rates

Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. Here, we are dealing with the rate at which the height of water in a cylindrical tank changes over time, related to the rate at which water enters the tank and the rate at which the radius of the cylinder increases.
These problems typically involve several steps:

• Identify and write down all known quantities and their rates of change.
• Express the relationship between these quantities using a formula.
• Differentiate this formula with respect to time.
• Substitute all known values and solve for the unknown rate of change.

This allows us to break down complex problems into manageable steps by connecting different moving parts through the application of calculus.
In the given problem, we connected the rate of water flow into the tank with the rates of change of the radius and height of the cylindrical tank to find the unknown rate at which the water height increases.

Differentiation with Respect to Time

To solve related rates problems, we need to use differentiation with respect to time. This process is called implicit differentiation. By doing this, we can find how one quantity changes over time concerning another.
The key formula in this problem is the volume formula of a cylinder, \( V = \pi r^2 h \). Differentiating this equation with respect to time \( t \), we use the chain rule which involves:\